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$L= \{a^{2^n} \mid n>1\} \cup \{a^m \mid m>1\}$

Is $L$ a regular?

My approach is-- union of a non regular and regular language may be regular or may be not regular. The left side of union is not regular so the language L is not regular.

Please advise.

J.-E. Pin
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Rahul
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  • The first set is a subset of the second one, isn't it? Why does this help you? – PhoemueX Nov 17 '15 at 09:46
  • so we can write the union as L={$a^m$,m>1}? can we say L is regular. – Rahul Nov 17 '15 at 09:56
  • Yes, you are right. I leave it to you to decide if ${a^m \mid m>1} $ is regular or not. – PhoemueX Nov 17 '15 at 10:01
  • Thanks for your help.The language L is regular,since we can draw a DFA for this. – Rahul Nov 17 '15 at 10:08
  • Your approach is bogus: you say that the union may be regular or not, then you immediately conclude that it is not. What's the logics ? –  Nov 17 '15 at 10:08
  • Hi Yves, I am trying to learn the concepts.Please advise which approach shall I try. – Rahul Nov 17 '15 at 11:10

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